Given y = (x^2+5x+7)/(x^2+5x+4) prove that y = 1/(x+1) - 1/(x+4).

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To prove that (x^2+5x+7)/(x^2+5x+4) =  1/(x+1) - 1/(x+4), we'll have to solve the difference between the 2 fractions from the right side.

We notice that the ratios don't have the same denominator, so we cannot compute the subtraction before calculating their LCD.

LCD = (x+1)*(x+4)

To obtain the same LCD to both ratios, we'll multiply the first ratio by x+4 and the second ratio by x+1.

 1/(x+1) - 1/(x+4) = [(x+4) - (x+1)]/(x+1)*(x+4)

We'll remove the brackets from numerator:

[(x+4) - (x+1)] = x + 4 - x - 1

We'll combine and eliminate like terms and we'll get:

[(x+4) - (x+1)] = 3

We'll multiply the brackets form denominator:

(x+1)*(x+4)  = x^2 + 4x + x + 4

We'll combine like terms and we'll get:

(x+1)*(x+4)  = x^2 + 5x + 4

 1/(x+1) - 1/(x+4) = 3/((x^2 + 5x + 4)

We remark that (x^2+5x+7)/(x^2+5x+4) = 3/((x^2 + 5x + 4) is a false statement. In this case, only the denominator is the same, but numerator is different.

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neela | High School Teacher | (Level 3) Valedictorian

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To prove that  if y = (x^2+5x+7), then y = 1/(x+1)-1/(x+4).

We see that the denominator is x^2+5c+4 = (x+1)(x+4).

We divide (x^2+5x+7)/(x^2+5x+4) =  1+ {(x^2+5x+7-(x^2+5x+4)}/(x^2+5x+4).

Also we notice that denominator x^2+5x+4 = (x+1)(x+4).

Therefore (x^2+5x+7)/(x^2+5x+4)  = 1+ 3/(x^2+5x+4).....(1)

We notice that 3 /(x^2+5x+4) = {(x+4)-(x+1)}/(x+1)(x+4)

3 /(x^2+5x+4) = 1/(x+1) - 1/(x+4). Substituting this in  (1), we get:

(x^2+5x+7)/(x^2+5x+4) = 1 + 1/(x+1)  - 1/(x+4).

Therefore if y = (x^2+5x+7)/(x^2+5x+4), then 

 y = 1 + 1/(x+1)- 1/(x+4) and not y = 1/(x+1) - 1/(x+4).

 

 

 

Therefore 3/(x+1)(x+4) = ((x+4)-(x+1))/(x

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